# C Program for Longest Palindromic Subsequence | DP-12

• Last Updated : 13 Aug, 2021

Given a sequence, find the length of the longest palindromic subsequence in it.

As another example, if the given sequence is “BBABCBCAB”, then the output should be 7 as “BABCBAB” is the longest palindromic subsequence in it. “BBBBB” and “BBCBB” are also palindromic subsequences of the given sequence, but not the longest ones.

1) Optimal Substructure:
Let X[0..n-1] be the input sequence of length n and L(0, n-1) be the length of the longest palindromic subsequence of X[0..n-1].
If last and first characters of X are same, then L(0, n-1) = L(1, n-2) + 2.
Else L(0, n-1) = MAX (L(1, n-1), L(0, n-2)).

Following is a general recursive solution with all cases handled.

## C

 `// C program of above approach``#include ``#include ` `// A utility function to get max of two integers``int` `max(``int` `x, ``int` `y) { ``return` `(x > y) ? x : y; }` `// Returns the length of the longest palindromic subsequence in seq``int` `lps(``char``* seq, ``int` `i, ``int` `j)``{``    ``// Base Case 1: If there is only 1 character``    ``if` `(i == j)``        ``return` `1;` `    ``// Base Case 2: If there are only 2 characters and both are same``    ``if` `(seq[i] == seq[j] && i + 1 == j)``        ``return` `2;` `    ``// If the first and last characters match``    ``if` `(seq[i] == seq[j])``        ``return` `lps(seq, i + 1, j - 1) + 2;` `    ``// If the first and last characters do not match``    ``return` `max(lps(seq, i, j - 1), lps(seq, i + 1, j));``}` `/* Driver program to test above functions */``int` `main()``{``    ``char` `seq[] = ``"GEEKSFORGEEKS"``;``    ``int` `n = ``strlen``(seq);``    ``printf``(``"The length of the LPS is %d"``, lps(seq, 0, n - 1));``    ``getchar``();``    ``return` `0;``}`
Output:
`The length of the LPS is 5`

Dynamic Programming Solution

## C

 `// A Dynamic Programming based C program for LPS problem``// Returns the length of the longest palindromic subsequence in seq``#include ``#include ` `// A utility function to get max of two integers``int` `max(``int` `x, ``int` `y) { ``return` `(x > y) ? x : y; }` `// Returns the length of the longest palindromic subsequence in seq``int` `lps(``char``* str)``{``    ``int` `n = ``strlen``(str);``    ``int` `i, j, cl;``    ``int` `L[n][n]; ``// Create a table to store results of subproblems` `    ``// Strings of length 1 are palindrome of length 1``    ``for` `(i = 0; i < n; i++)``        ``L[i][i] = 1;` `    ``// Build the table. Note that the lower diagonal values of table are``    ``// useless and not filled in the process. The values are filled in a``    ``// manner similar to Matrix Chain Multiplication DP solution (See``    ``// https:// www.geeksforgeeks.org/matrix-chain-multiplication-dp-8/). cl is length of``    ``// substring``    ``for` `(cl = 2; cl <= n; cl++) {``        ``for` `(i = 0; i < n - cl + 1; i++) {``            ``j = i + cl - 1;``            ``if` `(str[i] == str[j] && cl == 2)``                ``L[i][j] = 2;``            ``else` `if` `(str[i] == str[j])``                ``L[i][j] = L[i + 1][j - 1] + 2;``            ``else``                ``L[i][j] = max(L[i][j - 1], L[i + 1][j]);``        ``}``    ``}` `    ``return` `L[0][n - 1];``}` `/* Driver program to test above functions */``int` `main()``{``    ``char` `seq[] = ``"GEEKS FOR GEEKS"``;``    ``int` `n = ``strlen``(seq);``    ``printf``(``"The length of the LPS is %d"``, lps(seq));``    ``getchar``();``    ``return` `0;``}`
Output
`The length of the LPS is 7`

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